A car is driving directly north on the freeway at a speed of 115 km/h and a truck is leaving the freeway driving 86.6 km/h in a direction that is 35° west of north. Assume north to be in the +y-direction and east to be in the +x-direction. What is the magnitude of the velocity of the truck relative to the car?
A car is driving directly north on the freeway at a speed of 115 km/h and a truck is leaving the freeway driving 86.6 km/h in a direction that is 35° west of north. Assume north to be in the +y-direction and east to be in the +x-direction. What is the magnitude of the velocity of the truck relative to the car?
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![## Problem Statement
A car is driving directly north on the freeway at a speed of 115 km/h, and a truck is leaving the freeway driving 86.6 km/h in a direction that is 35° west of north. Assume north to be in the +y-direction and east to be in the +x-direction.
### Question:
What is the magnitude of the velocity of the truck relative to the car?
### Detailed Explanation:
1. **Car's Velocity:**
- The car is traveling directly north with a velocity \( \vec{v_c} \) of 115 km/h.
- Since north is defined in the +y-direction, we represent the car's velocity vector as:
\[
\vec{v_c} = 0\,\hat{i} + 115\,\hat{j} \, \text{km/h}
\]
where \( \hat{i} \) and \( \hat{j} \) are the unit vectors in the x (east) and y (north) directions respectively.
2. **Truck's Velocity:**
- The truck is moving 35° west of north with a velocity \( \vec{v_t} \) of 86.6 km/h.
- We need to resolve this velocity into its x and y components.
- The y-component (north component) is \( 86.6 \cos(35°) \).
- The x-component (west component, negative x-direction) is \( -86.6 \sin(35°) \).
Thus, the truck's velocity vector in component form is:
\[
\vec{v_t} = -86.6 \sin(35°)\,\hat{i} + 86.6 \cos(35°)\,\hat{j}
\]
3. **Relative Velocity:**
- The velocity of the truck relative to the car \( \vec{v_{t/c}} \) is given by the difference between the truck’s velocity and the car’s velocity:
\[
\vec{v_{t/c}} = \vec{v_t} - \vec{v_c}
\]
- Substituting the values:
\[
\vec{v_{t/c}} = (-86.6 \sin(35°)\,\hat{i} + 86.6 \cos(35°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff7f72a6-27ca-4fc6-a195-84c52d046220%2Fd309a576-6dbe-43e2-8442-b5e7cc8579e2%2F7g6id3j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement
A car is driving directly north on the freeway at a speed of 115 km/h, and a truck is leaving the freeway driving 86.6 km/h in a direction that is 35° west of north. Assume north to be in the +y-direction and east to be in the +x-direction.
### Question:
What is the magnitude of the velocity of the truck relative to the car?
### Detailed Explanation:
1. **Car's Velocity:**
- The car is traveling directly north with a velocity \( \vec{v_c} \) of 115 km/h.
- Since north is defined in the +y-direction, we represent the car's velocity vector as:
\[
\vec{v_c} = 0\,\hat{i} + 115\,\hat{j} \, \text{km/h}
\]
where \( \hat{i} \) and \( \hat{j} \) are the unit vectors in the x (east) and y (north) directions respectively.
2. **Truck's Velocity:**
- The truck is moving 35° west of north with a velocity \( \vec{v_t} \) of 86.6 km/h.
- We need to resolve this velocity into its x and y components.
- The y-component (north component) is \( 86.6 \cos(35°) \).
- The x-component (west component, negative x-direction) is \( -86.6 \sin(35°) \).
Thus, the truck's velocity vector in component form is:
\[
\vec{v_t} = -86.6 \sin(35°)\,\hat{i} + 86.6 \cos(35°)\,\hat{j}
\]
3. **Relative Velocity:**
- The velocity of the truck relative to the car \( \vec{v_{t/c}} \) is given by the difference between the truck’s velocity and the car’s velocity:
\[
\vec{v_{t/c}} = \vec{v_t} - \vec{v_c}
\]
- Substituting the values:
\[
\vec{v_{t/c}} = (-86.6 \sin(35°)\,\hat{i} + 86.6 \cos(35°
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